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The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is: (Specific heat of water liquid and water vapour are 4.2 kJ $K^{- 1} k g^{- 1}$ and $2.0 k J K^{- 1} k g^{- 1};$ heat of liquid fusion and vapourisation of water are $334 k J k g^{- 1}$ and $2491 k J k g^{- 1},$ respectively). $(\log 273 = 2.436, \log 373 = 2.572, \log 383 = 2.583)$

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$7.90 k J k g^{1} K^{- 1}$

$2.64 k J k g^{- 1} K^{- 1}$

$8.49 k J k g^{- 1} K^{- 1}$

$9.26 k J k g^{- 1} K^{- 1}$

answer is D.

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Detailed Solution

$\underset{273 K}{H_{2} O_{(s)}} \overset{Δ S_{1}}{\to} \underset{273 K}{H_{2} O_{(l)}} \overset{Δ S_{2}}{\to} \underset{373 K}{H_{2} O_{(l)}}$

$\underset{Δ S_{3}}{\to} \underset{373 K}{H_{2} O (g)} \underset{Δ S_{4}}{\to} \underset{383 K}{H_{2} O_{(g)}}$

$Δ S_{1} = \frac{Δ H + u s}{273} = \frac{334}{273} = 1.22$

$Δ S_{2} = m C G (\frac{T_{2}}{T_{1}}) = 4.2 \times C n (\frac{373}{273}) = 1.31$

$Δ S_{3} = \frac{Δ H_{v a p}}{373} = \frac{2491}{373} = 6.68$

$Δ S_{4} = m C G (\frac{T_{2}}{T_{1}}) = 2 C n (\frac{383}{373}) = 0.05$

Total entropy change $Δ S = 9.26 K J K g^{- 1} K^{- 1}$

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